Question

a chemist measures the energy change \\( \delta h \\) during the following reaction:

\\( 2 \mathrm { no } _ { 2 } ( g ) \
ightarrow \mathrm { n } _ { 2 } \mathrm { o } _ { 4 } ( g ) \quad \delta h = - 55.3 \mathrm { kj } \\)

use the information to answer the following questions.

this reaction is...	endothermic.
suppose \\( 47.7 \mathrm { g } \\) of \\( \mathrm { no } _ { 2 } \\) react.	yes, absorbed.
will any heat be released or absorbed?	yes, released.
	no.
if you said heat will be released or absorbed in the second part of this question, calculate how much heat will be released or absorbed.	\\( \square \mathrm { kj } \\)
be sure your answer has the correct number of significant digits.

Explanation:

Part 1: Determine if the reaction is endothermic or exothermic

In a chemical reaction, if the enthalpy change ($\Delta H$) is negative, the reaction releases heat, so it is exothermic. Here, $\Delta H = -55.3\ \text{kJ}$, so the reaction is exothermic.

Since the reaction is exothermic (releases heat) and $\ce{NO2}$ is a reactant, when $\ce{NO2}$ reacts, heat will be released.

Step 1: Find moles of $\ce{NO2}$

Molar mass of $\ce{NO2}$: $M = 14.01 + 2\times16.00 = 46.01\ \text{g/mol}$.
Moles of $\ce{NO2}$: $n = \frac{47.7\ \text{g}}{46.01\ \text{g/mol}} \approx 1.0367\ \text{mol}$.

Step 2: Relate moles to reaction enthalpy

The reaction is $2\ce{NO2}(g)
ightarrow \ce{N2O4}(g)$ with $\Delta H = -55.3\ \text{kJ}$.
For 2 moles of $\ce{NO2}$, heat released is $55.3\ \text{kJ}$.
For $n$ moles of $\ce{NO2}$, heat released ($q$) is:
$q = \frac{n}{2} \times 55.3\ \text{kJ}$.

Substitute $n = 1.0367\ \text{mol}$:
$q = \frac{1.0367}{2} \times 55.3 \approx 29.0\ \text{kJ}$.

Answer:

exothermic.

Part 2: Determine heat release/absorption for 47.7 g of $\ce{NO2}$